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Sagot :
To determine the interest earned by Mary and Josh during each of the first three years, we need to use the appropriate formulas for compound interest and simple interest respectively.
### Given Data:
- Principal [tex]\( P = \$30,000 \)[/tex]
- Interest Rate [tex]\( r = 4\% = 0.04 \)[/tex]
### Compound Interest (Mary)
The formula for compound interest when compounded annually is:
[tex]\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Since the interest is compounded annually, [tex]\( n = 1 \)[/tex], the formula simplifies to:
[tex]\[ A = P \times (1 + r)^t \][/tex]
### Simple Interest (Josh)
The formula for simple interest is:
[tex]\[ I = P \times r \times t \][/tex]
Now, let's calculate the interest for each of the first three years.
#### Year 1
Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^1 = 30,000 \times 1.04 = 31,200 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 31,200 - 30,000 = 1,200 \][/tex]
Josh:
[tex]\[ I = 30,000 \times 0.04 \times 1 = 1,200 \][/tex]
#### Year 2
Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^2 = 30,000 \times 1.0816 = 32,448 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 32,448 - 30,000 = 2,448 \][/tex]
Josh:
[tex]\[ I = 30,000 \times 0.04 \times 2 = 2,400 \][/tex]
#### Year 3
Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^3 = 30,000 \times 1.124864 = 33,745.92 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 33,745.92 - 30,000 = 3,745.92 \][/tex]
Josh:
[tex]\[ I = 30,000 \times 0.04 \times 3 = 3,600 \][/tex]
### Summary:
The interest earned by Mary and Josh during each of the first three years is summarized in the table below:
[tex]\[\begin{array}{|c|c|c|} \hline \text{Year} & \text{Interest Mary earns} & \text{Interest Josh earns} \\ \hline \text{First} & \$1,200 & \$1,200 \\ \hline \text{Second} & \$2,448 & \$2,400 \\ \hline \text{Third} & \$3,745.92 & \$3,600 \\ \hline \end{array}\][/tex]
### Conclusion:
Who earns more interest?
- In the first year, both Mary and Josh earn the same amount of interest: [tex]$1,200. - In the second and third years, Mary earns more interest than Josh. Specifically, in the second year she earns $[/tex]2,448 compared to Josh's [tex]$2,400, and in the third year, she earns $[/tex]3,745.92 compared to Josh's $3,600.
### Given Data:
- Principal [tex]\( P = \$30,000 \)[/tex]
- Interest Rate [tex]\( r = 4\% = 0.04 \)[/tex]
### Compound Interest (Mary)
The formula for compound interest when compounded annually is:
[tex]\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Since the interest is compounded annually, [tex]\( n = 1 \)[/tex], the formula simplifies to:
[tex]\[ A = P \times (1 + r)^t \][/tex]
### Simple Interest (Josh)
The formula for simple interest is:
[tex]\[ I = P \times r \times t \][/tex]
Now, let's calculate the interest for each of the first three years.
#### Year 1
Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^1 = 30,000 \times 1.04 = 31,200 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 31,200 - 30,000 = 1,200 \][/tex]
Josh:
[tex]\[ I = 30,000 \times 0.04 \times 1 = 1,200 \][/tex]
#### Year 2
Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^2 = 30,000 \times 1.0816 = 32,448 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 32,448 - 30,000 = 2,448 \][/tex]
Josh:
[tex]\[ I = 30,000 \times 0.04 \times 2 = 2,400 \][/tex]
#### Year 3
Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^3 = 30,000 \times 1.124864 = 33,745.92 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 33,745.92 - 30,000 = 3,745.92 \][/tex]
Josh:
[tex]\[ I = 30,000 \times 0.04 \times 3 = 3,600 \][/tex]
### Summary:
The interest earned by Mary and Josh during each of the first three years is summarized in the table below:
[tex]\[\begin{array}{|c|c|c|} \hline \text{Year} & \text{Interest Mary earns} & \text{Interest Josh earns} \\ \hline \text{First} & \$1,200 & \$1,200 \\ \hline \text{Second} & \$2,448 & \$2,400 \\ \hline \text{Third} & \$3,745.92 & \$3,600 \\ \hline \end{array}\][/tex]
### Conclusion:
Who earns more interest?
- In the first year, both Mary and Josh earn the same amount of interest: [tex]$1,200. - In the second and third years, Mary earns more interest than Josh. Specifically, in the second year she earns $[/tex]2,448 compared to Josh's [tex]$2,400, and in the third year, she earns $[/tex]3,745.92 compared to Josh's $3,600.
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